The X-rays produced in a Coolidge tube of pot | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The minimum wavelength (cutoff wavelength) of $X-$rays produced by an electron accelerated through a potential difference $V$ is given by the form

Vedclass · Accepted Answer

$3.09 	imes 10^{-11} \ m$

Vedclass · Answer

(A) The minimum wavelength (cutoff wavelength) of $X-$rays produced by an electron accelerated through a potential difference $V$ is given by the formula: $\lambda_{\min} = \frac{hc}{eV} = \frac{12400 \ \mathring{A} \cdot V}{V 	ext{ (in volts)}}$.Given $V = 40 \ kV = 40,000 \ V$.Substituting the values: $\lambda_{\min} = \frac{12400}{40000} \ \mathring{A} = 0.31 \ \mathring{A}$.Converting to meters: $0.31 	imes 10^{-10} \ m = 3.1 	imes 10^{-11} \ m$.Given the options provided,the closest value is $3.09 	imes 10^{-11} \ m$.

The $X-$rays produced in a Coolidge tube of potential difference $40 \ kV$ have a minimum wavelength of:

Similar Questions

$A$ potential difference of $42,000 \ V$ is used in an $X-ray$ tube to accelerate electrons. The maximum frequency of the $X-radiations$ produced is ($1 \ eV = 1.6 \times 10^{-19} \ J$ and $h = 6.63 \times 10^{-34} \ J \cdot s$).

An $X$-ray tube is operated at $50\, kV$. The minimum wavelength produced is ....... $\mathring{A}$.

When $X$-rays pass through a high-intensity uniform magnetic field,they:

If the potential difference $V$ applied to the Coolidge tube is doubled, then the cut-off wavelength

The electron transition required for the emission of characteristic $K_\beta$ $X$-rays is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module